H********g
发帖数: 43926 | 1 Muḥammad ibn Mūsā al-Khwārizmī[note 1] (Persian: مح&#
1605;د بن موسی خو&#
1575;رزمی, Arabic: مح&#
1605;د بن موسى ال&#
1582;وارزمی; c. 780
– c. 850) (Arabic pronunciation: [ælxɑːræzmiː]),
formerly Latinized as Algoritmi,[note 2]
a Persian[3] m... 阅读全帖 |
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r****a
发帖数: 1350 | 2
1. finding nontrivial upper and lower bounds of critical thresholds for
percolation on 3D lattice, and prove.
2. Prove the unsplittable multicommodity flow problem is NP-hard
3. With the additional constraint that for two given sets S1 and S2
of pairs of vertices, we need to ensure that every pair of vertices in S1 is
on the same side of the cut, while every pair of vertices in S2 is
separated by the cut, write new quadratic program and vector programs for
this constrained maximum cut problem a... 阅读全帖 |
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T*********n
发帖数: 1361 | 3 只知道怎么model一般问题成QP来进行优化，没写过
quadratic program and vector programs for
this constrained maximum cut problem
也许是这样做， 把 S1 ∩ S2 = ∅ , S1 ∪ S2 = S vectorize了写成constraint,
并且把 NetworkFlow(S1，S2) model出来，写成矩阵形式， maximize.
这个问题看起来像NP hard, 所以简单的approximation说不定也可以。。。
随便说说。。。
|
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p******e
发帖数: 17163 | 4 什么是一次二次方程？quadratic equation? |
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D*****r
发帖数: 6791 | 5 “The anatomical transition from reptiles to mammals is particularly well
documented in the key anatomical change of jaw articulation to hearing bones
. Only one bone, called the dentary, builds the mammalian jaw, while
reptiles retain several small bones in the rear portion of the jaw. We can
trace, through a lovely sequence of intermediates, the reduction of these
small reptilian bones, and their eventual disappearance or exclusion from
the jaw, including the remarkable passage of the reptilian... 阅读全帖 |
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J*****3
发帖数: 4298 | 6
论创造论者视而不见的进化过渡形态
“The anatomical transition from reptiles to mammals is particularly well
documented in the key anatomical change of jaw articulation to hearing bones
. Only one bone, called the dentary, builds the mammalian jaw, while
reptiles retain several small bones in the rear portion of the jaw. We can
trace, through a lovely sequence of intermediates, the reduction of these
small reptilian bones, and their eventual disappearance or exclusion from
the jaw, including the remarkable pas... 阅读全帖 |
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p***o
发帖数: 177 | 7 【 以下文字转载自 Physics 讨论区 】
【 原文由 susygut 所发表 】
我来补充补充物理的方面吧
说白了就是因为
1.我们不知道为什么有electroweak symmetry breaking
为什么是你说的100GeV
到底有没有Higgs，有了Higgs就有了quardratic divergence的问题
比如你算higgs的propogator，一圈有个top quark loop的修正
是cutoff^2发散的，（SM 其它的很多ln的）
susy的介绍，因为有了superpartner，是scalar正好和这个fermion的loop 正负号不同
所以消调了
你也可以算gauge loop这里也有gaugino的消除
所以susy可以说stabilize了higgs的质量
当然现在也有人做什么little higss企图解决这个问题
不过那个只能消除一圈的quadratic divergence
当然了就是在SUSY里
我们还是不知道为什么有个ELECTROWEAK SCALE
这就是大家说的MU PROBLEM
2.gauge unification
你 |
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f*****l
发帖数: 82 | 8 【 以下文字转载自 Statistics 讨论区 】
发信人: flywill (flywill), 信区: Statistics
标 题: 问概率问题
发信站: BBS 未名空间站 (Sat Feb 2 19:49:21 2008), 转信
X1, X2, X3...., Xn are n independent random variables, which are Gaussia
n distributed.
Define:
Y1 = a0+a1 X1+a2 X2 +...an Xn + a_(n+1)(X1)^2+ a_(n+2)(X2)^2+....+a_(2n)
(Xn)^2,
Y2 = b0+b1 X1+b2 X2 +...bn Xn + b_(n+1)(X1)^2+ b_(n+2)(X2)^2+....+b_(2n)
(Xn)^2,
如果才能快速得到max(Y1, Y2)的表达式，最好也写成X1到Xn的quadratic form，如
果能快速求出Pr(Y1 > ｃ)？ｃis a constant.
Many thanks:) |
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S*****T
发帖数: 400 | 9 好想不是很科普。呵呵
写这东西真是痛苦啊
根本不能想写什么当需要花大量时间考虑潜词造句是否正确的时候。
看来偶的中文真的退化了
sigh
超对称维像理论(Supersymmetry Phenomenology)
如前面所说超对称(SUSY)作为新物理的一个非常promising的候选。不仅为gauge
hierarchy 提供了一个非常elegant的解答而且也是一个自洽的超弦理论所要求。局域的
超对称理论也很自然的对应引力理论，被称为超引力。不过超对称尚未有直接的对撞机信
号作为证据。通常认为SUSY Gauge Unification是一个对超对称间接的提示。实验上对超
对称的直接搜索的努力从来也没有停止过。超对称同时又必须是一个破却的对称性因为我
们没有证据认为我们的自然是超对称的。超对称破却机制因此一直是一个理论热点。当然
超对称作为Gauge hierarchy问题的解要求超对称破却不能介绍进新的quadratic
divergence，必须是soft breaking而且SUSY partner的质量应该在TeV。所以超对称破却
的机制是受很强约束的。超对称在所谓的hidd |
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f*****p
发帖数: 235 | 10 【 以下文字转载自 EE 讨论区 】
【 原文由 firstep 所发表 】
Quadratic programming计算的是目标函数二次型，约束条件是线性等式或者
不等式。
如果反过来怎么计算？目标函数是线性函数，约束条件是二次型。多谢。 |
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k******y
发帖数: 17 | 11 more details here
1. if there is only one class, prove that L = average of all items
recall some knowledge on quadratic functions
2. for any two classes, x1 < x2 < .. < x_n and y1 < y2 < .. < y_n
there exists an optimal solution in which
there is no k s.t. x1 < y_k < x_n or y1 < x_k < y_n
that means any two classes should be disjoint in some sense
the following lemma may help you to prove this:
if x1 + .. + x_{n - 1} < y2 + .. + y_n, then x_n cannot > y1
because otherw |
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K****n
发帖数: 5970 | 12 i think neither linear nor quadratic is spatial invariant
your radial basis function looks spatially invariant to me |
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s***1
发帖数: 49 | 13 Why is the RBF spatially invariant but the linear and quadratic variant? |
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j*****t
发帖数: 9 | 14 Mixed integer linear programming
with linear objectives and quadratic constaints.
Thanks. |
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w***n
发帖数: 1084 | 15 Maybe the function can return the intersection point (P) between two
segments, such that:
|P-P1|==l1 and |P-P2|=l2,
which can be calculated from two quadratic equations with two unknowns. |
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f*********g
发帖数: 632 | 16 Princeton, 20 March 1956
Dear Mr. von Neumann:
GÄodel Book|Wigderson - rev. 2010-0708 5
With the greatest sorrow I have learned of your illness.
The news came to me as quite unexpected. Morgenstern
already last summer told me of a bout of weakness you once
had, but at that time he thought that this was not of any
greater significance. As I hear, in the last months you
have undergone a radical treatment and I am happy that this
treatment was successful as desired, and that you are now
doing ... 阅读全帖 |
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B****x
发帖数: 17 | 17 ACM Paris Kanellakis Theory and Practice Award奖励对计算机应用有重大影响的理论
发现. http://awards.acm.org/kanellakis/
2008
Cortes, Corinna
Vapnik, Vladimir
Paris Kanellakis Theory and Practice Award
2008 – Vladimir Vapnik
Fellow of NEC Laboratories/Columbia University (2008)
Citation
For the development of Support Vector Machines, a highly effective
algorithm for classification and related machine learning problems.
Full Citation
In their 1995 paper titled "Support vector networks," Cortes and Vapnik
introduced ... 阅读全帖 |
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i********c
发帖数: 54 | 18 Thanks Xentar for comments.
The original problem is simplified to some extent. In general, it might not be
quadratic objective function. It is the reason that I want to use GAMS, which
is supposed to be powerful for NLP, MINLP, etc.
However, it seems GAMS is quite limited on this type of problem ...
optimization
30,
bag
B, |
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c*****t
发帖数: 1879 | 19 统计有帮助，不过如果只是要学会用的话，有些基本概念就是了。就算学
统计，也不可能学得那么深入。这方面学精通非得多年的投入。如果只是
肤浅的学到怎么用，反倒是只要花几天囫囵吞枣的能大致明白就是了。
大致就是按这个顺序看看 tutorial 就是了。
1. linear discriminant / minimizing error / least square
2. quadratic discriminant
3. confusion matrix / accuracy / bias variance trade off etc
4. maximal margin classifier
5. SVM
6. kernel method
7. perceptron
8. neuro network
9. boosting
10. naive bayesian
11. decision tree / c4.5
12. bagging
13. random forest
14. voting / ensemble methods
15. PCA / dimension reduction et |
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H***a
发帖数: 735 | 20 经一位大侠指点，问题已经解决. 就是 lb 和 ub 也必须写成向量，自己汗一下
x = quadprog(tempV, -r', [], [], e', 1, [0 0 0 0], [1 1 1 1], [], options); |
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X****r
发帖数: 3557 | 21 There is really no point writing Python code like that.
If you like functional style, just use Haskell:
q s = case s of{[]->[];(x:xs)->q [y|y<-xs,y=x]}
Which does almost (except for the case the the pivot element is repeated)
exactly you do here and is more readable.
Note that this 'quicksort' (both the Python and the Haskell version)
is not really a quicksort, i.e. it does not sort in-place, thus has
quadratic space and time cost. |
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L***n
发帖数: 6727 | 22
取决与怎么model particle
我同意，我本来是认为对不同类型的函数,(linear, quadratic, ...)和不同的surface
union, 也许检查碰撞的impplementation需要不同，但是想想看，也许只是iterate是
否函数为0即可，这样你的设计确实更好。 |
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a*****e
发帖数: 1700 | 23 不是因为不让改值，是你还不太理解 laziness 和 data dependency。你看我写的就没
有内存问题。
如果你画一个图，每一行是数字 2^n 的数字列
然后下一行是根据上一行算出来的，如果把每个数字和计算它所需的上一行的数字之间
画一条线
那么你最后得到的是一个斜三角，计算最后一行的最后一个数字（如果按你的写法
most significant digit 在最后）所需要的 dependency 可以一直追溯到第一行第一
个数字。
所以这时候你用 lazy 的方式按需来算，就必须保证所需的 dependecies 在那里。这
就是为什么会消耗内存。
注意了 strictness 之后就没有内存消耗问题了，只是你这个算法本身是 quadratic，
所 n 越大越慢。 |
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h*******u
发帖数: 15326 | 24 感觉测了一个线性或quadratic solver. 这个太不准了，不同问题差大了 |
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a*****g
发帖数: 19398 | 25 #!/usr/bin/env pike
// legal.pike - Count the number of legal go boards.
// Copyright 2005 by Gunnar Farneb?ck
// [email protected]
/* */
//
// You are free to do whatever you want with this code.
//
//
// This program computes the number of legal go board configurations
// for a rectangular board of a given size. It is efficient enough to
// handle boards up to 8x8 within minutes and up to 11x11 in less than
// 24 hours (on a fast computer). For rectangular boa... 阅读全帖 |
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o*p
发帖数: 177 | 26 I would do:
repeated measures ANOVA
covariance matrix: AR1
test: group by time (linear or quadratic trend) interaction |
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y*****w
发帖数: 1350 | 27 What oop mentioned makes sense. It's a mixed model repeated measures
analysis.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
SAS code:
proc mixed data=data;
class group subject;
model y = group day group*day group*day*day / ddfm=kr;
repeated day / type=ar(1) subject=subject;
run;
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Note:
(1) Although ar(1) is supposed to be the best covariance structure, it would
be better to test other covariance structures ... 阅读全帖 |
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s***e
发帖数: 404 | 28 水里(加了很强的酸, PH=1左右)配的calibration curve, 是开口向上的quadratic曲线
, 以前从来没出现过这样的情况 (都是直线). 有没有人知道是怎么回事啊? 怎样解决
这个问题? 而且8个分析物sensitivity也不稳, 一个样品连续进样两次,每个化合物的
intensity都会变, 而且变化方向不一样, 比方说第一次进样, A的cps达到3000, B达到
2000, 第二次就可能是A最高打到1500, B变成3000. Mass spec清理了几遍, 还是这样. 曲线就象这样: |
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a***n
发帖数: 1152 | 29 quadratic curve也是curve啊。为什么不能用？
你的chromatogram怎样？ Is retention time consistent?
样. 曲线就象这样: |
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w**d
发帖数: 2334 | 30 No idea why bilinear is better than linear.
Long time ago, by using an unusual strategy,
we obtained a O(h^2) rectangular element
with 12 degree of freedom. The error is a little bigger than that from
bi-quadratic basis. But the order is the same.
I know nothing about superconvergence.
The high convergence rate might be due to some special property of
the conformation tensor or even your mesh.
BTW, i applied for the postdoc position at Rice. Not sure if i can get it. |
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l***e
发帖数: 10 | 31 Hi, All:
I came up with an optimization problem min f(x)= 1/2 x' H x- abs(x) subject
to -c<=x<=c, and sum(x)=0. Here, H is a matrix, x is vector, c is bound.
It looks like quadratic programming. But not really.
I tried the "fmincon" in the optimzation toolbox of Matlab, but it can't
find a statisfactory solution. The problem is not convex ( I think), but
some optimization method should reach a good minima.
Anyone can help? Please! If anyone can solve this, it could be used in a
wonderful |
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e***n
发帖数: 286 | 32 basically, it seems to me the newton method is among the fastest quality-
guaranteed algorithms. It is quadratic covergence. All other improvements
are at most superlinear convergence, such as various quasi-newton methods.
but newton methods are rarely used to solve practical problems, since it is
not descent-guaranteed and it consumes memory a lot. |
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S*M
发帖数: 536 | 33 cplex只能处理quadratic，对于一般的非线性问题没办法吧
而且cplex也不便宜 |
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l*****i
发帖数: 3929 | 34 google "OOQP", written by Steve Wright and Mike Gertz |
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H***a
发帖数: 735 | 35 经一位大侠指点，问题已经解决. 就是 lb 和 ub 也必须写成向量，自己汗一下
x = quadprog(tempV, -r', [], [], e', 1, [0 0 0 0], [1 1 1 1], [], options); |
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j*******o
发帖数: 246 | 36 体积小一点，移植方便点。小弟EE出身，编程水平是在一般。。。太大太复杂的是在看
不懂。。。 |
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t***s
发帖数: 4666 | 40 it may not be linear, quadratic, or even convex. |
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j**u
发帖数: 6059 | 41 ☆─────────────────────────────────────☆
cityhawk (呆鹰) 于 (Mon May 23 20:38:14 2011, 美东) 提到:
Matlab程序是 for 嵌套循环：比如，
a=0.1:0.5 with spacing 0.01; b=0.1:0.6 with spacing 0.01
c=0.1:0.8 with spacing 0.01; d=0.1:0.6 with spacing 0.01
e=0.1:0.9 with spacing 0.01; f=0.1:0.7 with spacing 0.01
g=0.1:0.6 with spacing 0.01; h=0.1:0.5 with spacing 0.01
执行部分
end; end; end; end;end; end; end; end;
这个程序在普通的PC 3.6GHz, 2GB内存上运行要2个星期多，把它放在系里的服务器上
运行，结果比我们lab的这个PC还慢，网管告诉我系里服务器的单个CPU才1.8GHz，尽管
我们有近30个CPU并行和全部 2... 阅读全帖 |
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j******a
发帖数: 1599 | 42 我们学校的数据库没有老的paper。那位能帮个忙，发到j******[email protected]
包子答谢了。
International Journal of Control, Volume 48, Issue 2 August 1988 , pages 481
- 498
Controllability, observability and discrete-time markovian jump linear
quadratic control
Authors: Yuandong Ji a; Howard J. Chizeck a |
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e******i
发帖数: 179 | 43 高斯矩阵quadratic expectation
能展开吗？
A,B,C是三个随机矩阵
E(ABC)能展开吗？ |
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j*****t
发帖数: 9 | 45 Mixed integer linear programming
with linear objectives and quadratic constaints.
Any suggestion?
Thanks. |
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z*b
发帖数: 12 | 46 yes, you need to approximate your curve with some smooth curve before yo
u do , for example quadratic functions (that's simpson's rule), but of c
ourse the accuracy might not be good enough, so you may use piecewise qu
adratic functions or in general consider spline functions. Another techn
ique called least squares fitting might also give a good smooth curve fo
r you to integrate using standard quadrature formulas.
numerical
looks |
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k*******l
发帖数: 69 | 47 is this a very 'deep' subject?
there are numerous standard solvers on the net
NO need to research on this at all |
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c**c
发帖数: 39 | 48 S is the sample covariance matrix of a random vector X.
From singular value decomposition,
S=UDU'
Partition the columns of U as (U1,U2).
W is the sample covariance matrix of vec(S), that is, the sample fourth-
moments of the random vector X.
The singular value decomposition of W is
W=AGA'
Hence, the inverse of W is Winv= A(G^-1)A'.
Define
kx as kronecker product,
and define two quadratic forms
L1=y'(U2'kxU2')Winv(U2kxU2)y = y'(U2'kxU2')A(G^-1)A'(U2kxU2)y
L2=y'(U2'kxU2')(G^-1)(U2kxU2)y
The questi |
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k*******l
发帖数: 69 | 49 可以不独立啊，quadratic covariation不为0就行了
rho(1, 2)未必一定是常数 |
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k*******l
发帖数: 69 | 50 ft
the quadratic covariation of two BMs is not necessarily rho*t
this is just model assumption, since time-dependent or stochastic
instantaneous correlation (the rho above) is too hard to estimate/calibrate
rate |
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